\[A=9 \sqrt{3}=\frac{1}{2} b h\]
So we have that the base, b, of the triangle is s
b=s
So we have
\[9 \sqrt{3}=\frac{1}{2}s h\]
So this means we have
\[(2) 9 \sqrt{3} =\frac{1}{2}sh (2)\]
\[18 \sqrt{3} =sh\]
So is there a way to write h in terms of s
Well remember we have an equilateral triangle with each side s
and h, the height of the equilateral can be found in terms of s by using the Pythagorean Theorem
|dw:1337557593984:dw|

\[h^2=s^2-\frac{s^2}{4}\]
Combine the fractions on the right hand side of the equation
\[h^2=\frac{4s^2}{4}-\frac{s^2}{4} \]
\[h^2=\frac{3s^2}{4}\]
Take square root of both sides
So we have
\[h=\sqrt{\frac{3s^2}4}\]
\[h=\frac{\sqrt{3s^2}}{\sqrt{4}}\]
\[h=\frac{\sqrt{3} \sqrt{s^2}}{2}\]
\[h=\frac{\sqrt{3}s}{2}\]